Step-by-step Solution:

1. Analyze the complex number z: The given complex number is $z = \frac{{ - 1 + \sqrt 3 i}}{2}$. This is a standard complex number, which can be expressed in polar form as $\cos(2\pi/3) + i\sin(2\pi/3) = e^{i2\pi/3}$. This is one of the complex cube roots of unity, commonly denoted by $\omega$. So, $z = \omega$. The key properties of $\omega$ are:

   • $\omega^3 = 1$
   • $1 + \omega + \omega^2 = 0$

2. Define the matrix P and calculate P²: The matrix P is given by: $P = \left[ {\begin{matrix} {{{( - z)}^r}} & {{z^{2s}}} \\ {{z^{2s}}} & {{z^r}} \\ \end{matrix} } \right]$ Let's compute $P^2$: $P^2 = P \cdot P = \left[ {\begin{matrix} {{{( - z)}^r}} & {{z^{2s}}} \\ {{z^{2s}}} & {{z^r}} \\ \end{matrix} } \right] \left[ {\begin{matrix} {{{( - z)}^r}} & {{z^{2s}}} \\ {{z^{2s}}} & {{z^r}} \\ \end{matrix} } \right]$ P^2 = \left[ {\begin{matrix} {{{(( - z)}^r)}^2} + {{(z^{2s})}^2}} & {{{( - z)}^r}{z^{2s}} + {z^{2s}}{z^r}} \\ {{z^{2s}}{{( - z)}^r} + {z^r}{z^{2s}}} & {{{(z^{2s})}^2} + {{(z^r)}^2}} \\ \end{matrix} } \right] Simplifying the terms:

   • Top-left and bottom-right elements are equal: $((-z)^r)^2 + (z^{2s})^2 = (-1)^{2r}z^{2r} + z^{4s} = z^{2r} + z^{4s}$
   • Off-diagonal elements are equal: $( - z)^r z^{2s} + z^{2s} z^r = z^{2s}((-1)^r z^r + z^r) = z^{2s} z^r ((-1)^r + 1)$ So, $P^2 = \left[ {\begin{matrix} {z^{2r} + z^{4s}} & {z^{2s+r}((-1)^r + 1)} \\ {z^{2s+r}((-1)^r + 1)} & {z^{2r} + z^{4s}} \\ \end{matrix} } \right]$

3. Apply the condition P² = -I: We are given $P^2 = -I$, where $I = \left[ {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right]$. $P^2 = \left[ {\begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} } \right]$ Comparing the elements of the matrices, we get two equations: (i) $z^{2r} + z^{4s} = -1$ (ii) $z^{2s+r}((-1)^r + 1) = 0$

4. Solve the system of equations: From equation (ii), since $z \neq 0$, we must have $z^{2s+r} \neq 0$. Therefore, the other factor must be zero: $(-1)^r + 1 = 0$ $(-1)^r = -1$ This implies that $r$ must be an odd integer. Since $r \in \{1, 2, 3\}$, the possible values for $r$ are $r=1$ and $r=3$.

   Now we use equation (i) and substitute $z=\omega$: $\omega^{2r} + \omega^{4s} = -1$ From the property $1 + \omega + \omega^2 = 0$, we know that $\omega + \omega^2 = -1$. This means that for the equation $\omega^A + \omega^B = -1$ to hold, the set of powers {A mod 3, B mod 3} must be {1, 2}.

   Let's check the possible values of $r$:

   Case 1: r = 1 The equation becomes $\omega^{2(1)} + \omega^{4s} = -1$, which is $\omega^2 + \omega^{4s} = -1$. For this to be true, we need $\omega^{4s} = \omega$. This means the exponent $4s$ must be congruent to 1 modulo 3: $4s \equiv 1 \pmod 3$ Since $4 \equiv 1 \pmod 3$, this simplifies to $s \equiv 1 \pmod 3$. Given that $s \in \{1, 2, 3\}$, the only value that satisfies this condition is $s=1$. So, $(r, s) = (1, 1)$ is a solution.

   Case 2: r = 3 The equation becomes $\omega^{2(3)} + \omega^{4s} = -1$, which is $\omega^6 + \omega^{4s} = -1$. Since $\omega^3 = 1$, we have $\omega^6 = (\omega^3)^2 = 1^2 = 1$. The equation becomes $1 + \omega^{4s} = -1$, which simplifies to $\omega^{4s} = -2$. This is impossible, because any power of $\omega$ lies on the unit circle in the complex plane, so its magnitude is 1. i.e., $|\omega^{4s}| = |\omega|^{4s} = 1^{4s} = 1$. However, $|-2| = 2$. Therefore, there are no solutions when $r=3$.

5. Conclusion: The only ordered pair $(r, s)$ that satisfies the given condition is $(1, 1)$. Thus, the total number of such ordered pairs is 1.